Final answer:
The rate at which the height of the pile is increasing can be found using the concept of related rates. By relating the volume of the cone to the rates of change of the height and radius, we can derive an equation to solve for the rate of change of height. Substituting the given values, the height of the pile is increasing at a rate of approximately 0.457 ft/min when the pile is 10 ft high.
Step-by-step explanation:
To find the rate at which the height of the pile is increasing, we can use the concept of related rates. Let's denote the height of the pile as h and the radius of the base as r. Since the base diameter and height are always equal, the radius is equal to half the height, so r = h/2.
The volume of a cone can be calculated using the formula V = (1/3) * π * r^2 * h. We are given that the gravel is being dumped at a rate of 30 ft^3/min, so the rate of change of the volume of the cone is 30 ft^3/min.
Using the formula for the volume of a cone, we can express the rate of change of the volume in terms of the rates of change of the height and radius. Taking the derivative with respect to time, we get dV/dt = (1/3) * π * (2rh * dh/dt + r^2 * dh/dt). Rearranging the equation, we can solve for dh/dt:
30 = (1/3) * π * (2rh * dh/dt + r^2 * dh/dt)
Plugging in the values for r and h, we have:
30 = (1/3) * π * (2(h/2)(dh/dt) + (h/2)^2 * dh/dt)
Simplifying the equation, we get:
30 = (1/3) * π * (h * dh/dt + (h^2)/4 * dh/dt)
Combining like terms, we have:
30 = (1/3) * π * ((5h^2)/4 * dh/dt)
Now, we can solve for dh/dt:
dh/dt = (30 * 4 * 3) / (π * 5h^2)
Substituting h = 10 ft, we get:
dh/dt = (30 * 4 * 3) / (π * 5(10^2))
dh/dt = (360 / (50π)) ft/min
Therefore, the height of the pile is increasing at a rate of approximately 0.457 ft/min when the pile is 10 ft high.